Geometry help needed

Question

Rectangle ABCD is symmetric with respect to y-axis. Points A and B belong to the parabola y=x². Points C and D are on the parabola y=−3x²+k. Find k if the area of the rectangle is 66 and length of AB is 6.

Answer 1

Hi Jack K.,

For y = x², the parabola opens upward and the vertex is at (0, 0).

For y = -3x² + k, the parabola opens downward with the vertex (0, k).

For the rectangle, A = L*W or L = A/W, so L = 66/6 = 11. So the rectangle has a width of 6 in the x-direction and the length of 11 in the y-direction. The rectangle is symmetric along the y-axis so the rectangle has x-values of -3 and 3.

If we use these x-values in y = x², then y = 9, and we see that A & B have coordinates of (-3, 9) and (3, 9).

With a length of 11 for the rectangle, we can plot two different rectangles with C & D coordinates at (-3, 20) and (3, 20) [y = 9 + 11], or with C & D at (-3, -2) and (3, -2) [y = 9 - 11].

So we get one parabola at:

y = -3x² + k

20 = -3(3)² + k

47 = k

y = -3x² + 47

And the other at:

y = -3x² + k

-2 = -3(3)² + k

25 = k

y = -3x² + 25

I hope this helps, Joe.